countable additivity for a collection of algebra imply countable subadditivity hi all does countable additivity for a collection of algebra imply countable subadditivity for an Algebra? if so, what is so special about being Algebra that makes countable subadditivity automatically holds when countable additivity holds?
thank you
 A: So, trying to make sense of this question, suppose we have $\mathcal{A}$ which is an algebra of subsets of some set $X$, and a countably additive function $\mu: \mathcal{A} \to [0,+\infty]$, say.
Then indeed $\mu$ is countably subadditive:
let $A_n, n \in \mathbb{N}$ be any family of sets from $\mathcal{A}$ such that $\bigcup_n A_n \in \mathcal{A}$.
Then define $B_0 = A_0$, $B_n = A_n \setminus \cup_{i=0}^{n-1} A_i$ for $n \ge 1$. The fact that $\mathcal{A}$ is an algebra (!!) means that all $B_n \in \mathcal{A}$ as well (algebras are closed under finite unions, intersections and set differences). 
Also, if $n \neq m$: $B_n \cap B_m = \emptyset$ (suppose $n < m$ WLOG, then $x \in B_n \cap B_m$ implies $x \in B_n$ so $x \in A_n$ and $x \in B_m$ so $x \notin A_n$ (it's one of the sets we substract from $A_m$ to form $B_m$!), contradiction, so the intersection is empty). So the $B_n$ form a pairwise dijsoint family , which is part one of their raison d'être. Part two is:
$$\bigcup_n B_n = \bigcup_n A_n$$ 
The left to right inclusion is clear, as $B_k \subseteq A_k$ for all $k$, So take $x \in \bigcup_n A_n$, and let $m$ be the minimal index in $\mathbb{N}$ such that $x \in A_m$ (there exists at least one such index, so there is a smallest one). Then $x \notin A_i$ for $i < m$ by minimality so either $x \in A_0 = B_0$ (if $m=0$) or $x \in A_m \setminus \cup_{i=1}^{m-1} A_i = B_m$ otherwise. So $x \in \bigcup_n B_n$ and we have equality.
Now the fact that $\mu$ is countably additive gives the middle equality in
$$\mu(\bigcup_n A_n) = \mu(\bigcup_n B_n) = \sum_n \mu(B_n) \le \sum_n \mu(A_n)$$
where the last inequality follows from $B_n \subseteq A_n$ so $\mu(B_n) \le \mu(A_n)$.
See how we use the algebra properties of $\mathcal{A}$ to construct the $B_n$? They have to be in $\mathcal{A}$ or $\mu$ would not be defined on them.
